# If $\{u,v,w\}$ is a basis for $V$, then $\{u+v+w,v+w,w\}$ is also a basis: is this proof correct?

Let $u,v,w \in V$ a vector space over a field F such that $u \neq v \neq w$. If $\{ u , v , w \}$ is a basis for $V$, then prove that $\{ u+v+w , v+w , w \}$ is also a basis for $V$.

Proof:

Let $u,v,w \in V$ a vector space over a field $F$ such that $u \neq v \neq w$. Let $\{ u , v , w \}$ be a basis for $V$. Because $\{ u , v , w \}$ is a basis, then $u,v,w$ are linearly independent and $\langle \{ u , v , w \} \rangle = V$.

Let $x \in V$ be an arbitrary vector then $x$ can be uniquely expressed as a linear combination of $\{ u , v , w \}$. Let's suppose $x=au+bv+cw$ for some $a,b,c \in F$.

On the other hand, let us consider $\{ u+v+w , v+w , w \} \subseteq V$.

Then \begin{align*} \langle \{ u+v+w , v+w , w \} \rangle &= \{d(u+v+w) + e(v+w) + f(w) \mid d,e,f \in F\} \\ &= \{du + (d+e)v +(d+e+f)w \mid d,e,f \in F \} . \end{align*}

If $x \in V$, then $x=du + (d+e)v +(d+e+f)w$ is another unique representation of $x \in V$ . Then for any arbitrary $x \in V$, we have $d=a$, $d+e=b$and $d+e+f=c \in F$.

Because $\{ u , v , w \}$ is a basis for $V$, then $\{ u+v+w , v+w , w \}$ must also be a basis for $V$.

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I tried to give an alternate proof instead of proving <{ u , v , w }>=V<{ u+v+w , v+w , w }> – aortizmena Dec 6 '11 at 5:39
Note that $u\neq v\neq w$ only means that $u\neq v$ and $v\neq w$, and does not mean that also $u\neq w$ ("is not equal to" is not transitive). Note also that saying that the vectors are unequal is redundant when you assume they form a basis. No two elements of a linearly independent set can be equal. – Jonas Meyer Dec 6 '11 at 5:41
Thanks. I typed that because that is how the statement was given. I do undestand that no two elements of a linearly independent set can be equal. – aortizmena Dec 6 '11 at 5:48
Are you aware of the fact that you can type LaTeX commands for your math notation? – user13838 Dec 6 '11 at 5:49
I should say that you are very, very close to proving that they do span. You just need to take in an arbitrary triple $a, b, c$ and produce $d, e, f$ such that the equations in your penultimate paragraph hold. – Dylan Moreland Dec 6 '11 at 6:00

In general I find it much harder to show that a set of vectors spans the vector space, than showing a set is independent. If you want to go about your approach and show they span, what you would need to do is take an arbitrary vector in $V$ and write it is a linear combination of your new set.

An alternative way is to show that they are independent, which turns out to be quite simple.

Suppose $$c_1(u + v + w) + c_2(v+w) + c_3(w) = 0$$ then we know that $$c_1u + (c_1 + c_2)v + (c_1 + c_2 + c_3) w = 0$$ but $u, v, w$ are independent and so $$c_1 = c_1 + c_2 = c_1 + c_2 + c_3 = 0$$ From here its pretty clear that $c_1 = c_2 = c_3 = 0$ which would prove the claim.

EDIT: Note, if we show that the vectors are independent they must also span $V$, because in any vector space of dimension $n$, if we have $n$ independent vectors, they must span the vector space. Similarly, if we were able to show that they spanned the space then they would have to be independent since we have $\dim(V)$ of them

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Showing they span isn't too bad either, given that $v=(v+w)-w$ and $u=(u+v+w)-(v+w)$. – Jonas Meyer Dec 6 '11 at 5:59
To add to @Jonas' comment: To prove that any set $S$ spans $V$, it suffices to show that $S$ spans $u$, $v$ and $w$. One need not explicitly worry about an arbitrary $x$ in $V$. – Srivatsan Dec 6 '11 at 6:01
Thanks. Does that also prove that the new basis spans V? – aortizmena Dec 6 '11 at 6:08
@Andres yes it does, see the edit – Deven Ware Dec 6 '11 at 6:12
Nice alternative – aortizmena Dec 6 '11 at 6:20

The proof is essentially correct, but you do have some unnecessary details. Removing redundant information, we can reduce it to the following:

Let $V$ be a vector space over $F$ and $\{u, v, w\}$ a basis for $V$. Any $x \in V$ can be uniquely written $x = au + bv + cw$ for some $a, b, c \in F$. Let $d, e, f \in F$ be the unique choices so that $d = a$, $e = b - a$, and $f = c - b$; then $x$ can be uniquely written as $x = du + (d + e)v + (d + e + f)w$. Rearranging, we have $x = d(u + v + w) + e(v + w) + fw$. Since this is a unique expression for an arbitrary $x \in V$, then $\{u + v + w, v + w, w\}$ is a basis for $V$.

There is also this nice alternate proof:

Let $V$ be a vector space over $F$ and $\{u, v, w\}$ a basis for $V$. Let's start a new basis with $w$. Since $\{u, v, w\}$ is linearly independent, $v + w \notin span\{w\}$ and $u + v + w \notin span\{w, v + w\}$, so $\{w, v + w, u + v + w\}$ is linearly independent. It is also the same length as a basis, so it must be a basis itself.

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I do not follow your proof, but I suspect that it is incomplete. You demonstrate that $x$ can be written as a linear combination of the vectors $\{ u+v+w, v+w, v \}$, and I follow this part. But I do not see where you prove the uniqueness of this representation. – Srivatsan Dec 6 '11 at 6:05
the uniqueness follows because the sets $\{u+v+w, v+w, v\}$ and $\{u,v,w\}$ have the same cardinality, no? – user12014 Dec 6 '11 at 6:09
It all depends on the theorems you have, I guess. – Dylan Moreland Dec 6 '11 at 6:10
Isn't saying that for for any arbitrary x∈V=<{ u , v , w }> with d=a, d+e=b and d+e+f=c it implies the unique solution d=a, e=b-d=b-a , f=c-d-e=c-a-(b-a)=c-a-b+a=c-b ? – aortizmena Dec 6 '11 at 6:10
since $x = au + bv + cw$ is unique and each $(a, b, c)$ corresponds to exactly one $(d, e, f)$, then $x = d(u + v + w) + e(v + w) + fw$ must also be unique. – smackcrane Dec 6 '11 at 6:12

It is easily seen that each of your original basis elements are in the span of $\{u+v+w,v+w,w\}$. Since $\{u, v, w\}$ spans $V$, so does $\{u+v+w,v+w,w\}$ (if $A\subseteq {\rm span} \,(V)$, then ${\rm span}(A)\subseteq{\rm span }(V) \thinspace$).

So, we have a set of three vectors that span a space of dimension three. It follows that the set is independent and thus a basis.

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